A resistor of `80 Omega` in a current element X is connected across an AC supply of 150 V. Calculate the rms value of current in the circuit if current is ahead of voltage in phase by `45^(@)` . Also identify the element X.

Answer

To solve the problem step by step, we will follow the outlined procedure:

### Step 1: Identify the given values
- Resistance (R) = 80 Ω
- Voltage (V_rms) = 150 V
- Phase angle (φ) = 45°

### Step 2: Determine the type of circuit element X
Since the current is ahead of the voltage by 45°, this indicates that the circuit has a capacitive nature. Therefore, element X is a capacitor.

### Step 3: Relate the impedance to the resistance and phase angle
In an AC circuit, the relationship between the resistance (R), impedance (Z), and phase angle (φ) is given by:
\[ Z \cos(φ) = R \]
Where:
- \( φ = 45° \)

### Step 4: Calculate the impedance (Z)
Using the cosine of 45°:
\[ \cos(45°) = \frac{1}{\sqrt{2}} \]
Substituting the values into the equation:
\[ Z \cdot \frac{1}{\sqrt{2}} = 80 \]
Rearranging to find Z:
\[ Z = 80 \cdot \sqrt{2} \]
Calculating \( Z \):
\[ Z \approx 80 \cdot 1.414 \approx 113.14 \, \Omega \]

### Step 5: Calculate the RMS value of the current (I_rms)
Using Ohm's law for AC circuits:
\[ I_{rms} = \frac{V_{rms}}{Z} \]
Substituting the known values:
\[ I_{rms} = \frac{150}{113.14} \]
Calculating \( I_{rms} \):
\[ I_{rms} \approx 1.32 \, A \]

### Final Answer
The RMS value of the current in the circuit is approximately **1.32 A**, and the element X is a **capacitor**.

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